Question

Find angle between unit vectors $\hat i + \hat j + \hat k$ and $\hat i + \hat j$ by cross product?

Answer 1

Here, we have to find the angle between unit vectors $\hat i + \hat j + \hat k$ and $\hat i + \hat j$ by cross product. Cross product of two vectors can be defined as a binary operation on two vectors in three dimensional spaces and denoted by $\times$ . In order to find the angle between given vectors by using cross product we use the formula which is $\left| {\vec c} \right| = \left| a \right|\left| b \right|\sin \theta$ where $a$ and $b$ are the magnitudes of the vector, $c$ is the magnitude of the vector product and $\theta$ is the angle between these two vectors.

Complete answer:
The cross- vector product, area product, or the vector product of two vectors can be defined as a binary operation on two vectors in three dimensional spaces and denoted by $\times$ .
The magnitude of the vector product can be given as $\left| {\vec c} \right| = \left| a \right|\left| b \right|\sin \theta$ , where $a$ and $b$ are the magnitudes of the vector, $c$ is the magnitude of the vector product and $\theta$ is the angle between these two vectors.
Here we have to find the angle between two unit vectors. So,
$\sin \theta = \dfrac{{\left| {\vec c} \right|}}{{\left| {\vec a} \right|\left| {\vec b} \right|}}$
Let $\left| {\vec a} \right| = \hat i + \hat j + \hat k$ and $\left| {\vec b} \right| = \hat i + \hat j$
Now, we will find the magnitude of the vectors. We have,
$\Rightarrow \left| {\vec a} \right| = \sqrt {{{(1)}^2} + {{(1)}^2} + {{(1)}^2}}$
Solving the square root. We get,
$\Rightarrow \left| {\vec a} \right| = \sqrt 3$
$\Rightarrow \left| {\vec b} \right| = \sqrt {{{(1)}^2} + {{(1)}^2}}$
Solving the square root. We get,
$\Rightarrow \left| {\vec b} \right| = \sqrt 2$
Therefore, the magnitude of $\left| {\vec a} \right| = \sqrt 3$ and $\left| {\vec b} \right| = \sqrt 2$
Now, we will calculate the magnitude of the cross product for that we have to find the cross product of two vectors. So,
$\Rightarrow \vec a \times \vec b = (\vec i + \vec j + \vec k) \times (\vec i + \vec j)$
$\Rightarrow \vec a \times \vec b = (\hat i \times \hat i) + (\hat i \times \hat j) + (\hat j \times \hat i) + (\hat j \times \hat j) + (\hat k \times \hat i) + (\hat k \times \hat j)$ .
We know that in cross product $\hat i \times \hat i = \hat j \times \hat j = \hat k \times \hat k = 0$ and $\hat i \times \hat j = \hat k,\,\,\,\hat j \times \hat i = - \hat k,\,\,\,\hat k \times \hat i = \hat j,\,\,\,\hat k \times \hat j = - \hat i$
Therefore,
$\Rightarrow \vec a \times \vec b = 0 + \hat k + ( - \hat k) + 0 + \hat j + ( - \hat i)$
Cancelling out the equal term with the opposite sign. We get,
$\Rightarrow \vec a \times \vec b = \hat j - \hat i$
Now, the magnitude of the cross product will be
$\Rightarrow \left| {\vec a \times \vec b} \right| = \sqrt {{{(1)}^2} + {{(1)}^2}}$
Solving the square root. We get,
$\Rightarrow \left| {\vec a \times \vec b} \right| = \sqrt 2$
Now, angle between vector is given by $\sin \theta = \dfrac{{\left| {\vec c} \right|}}{{\left| {\vec a} \right|\left| {\vec b} \right|}}$
Therefore,
$\Rightarrow \sin \theta = \left[ {\dfrac{{\sqrt 2 }}{{\sqrt 3 \cdot \sqrt 2 }}} \right]$
Cancelling out the equal terms. We get,
$\Rightarrow \sin \theta = \left[ {\dfrac{1}{{\sqrt 3 }}} \right]$
Taking the inverse of $\sin$ function. we get,
$\Rightarrow \theta = {\sin ^{ - 1}}\left[ {\dfrac{1}{{\sqrt 3 }}} \right]$
$\Rightarrow \theta = 35.26^\circ$
Hence, the angle between unit vectors $\hat i + \hat j + \hat k$ and $\hat i + \hat j$ is $35.26^\circ$ .

Note:
The vector product of two vectors basically refers to a vector that is perpendicular to both of the vectors and can be obtained by multiplying their magnitudes by the $\sin$ of the angle that exists between them. We can calculate the direction of the vector product with the help of the right- hand thumb rule in which we curl our fingers of the right hand around a line perpendicular to the plane of vectors $a$ and $b$ , then stretched thumb points in the direction of $c$ .